1. 1443-501 Spring 2002 Lecture #10 Dr. Jaehoon Yu 1.Term exam results 2.Linear Momentum 3.Momentum Conservation 4.Impulse and Momentum 5.What are Collisions? 6.Elastic and Inelastic collisions 7.Two dimensional collisions Homework: http://hw.utexas.edu/studentInstructions.html – Do Homework #2.http://hw.utexas.edu/studentInstructions.html Term Exam Answer Key: A new link on my web page http://www-hep.uta.edu/~yuhttp://www-hep.uta.edu/~yu

2. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 2 This mean value is what I am going to use to scale!! Some of you have done rather well. But most of you have done not as well as I was hoping for. What I expected was about twice as much as what your mean value is. Please do homework for extra credit and for yourselves!! Might introduce pop- quiz after mid-term.

3. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 3 Linear Momentum The principle of energy conservation can be used to solve problems that are harder to solve just using Newton’s laws. It is used to describe motion of an object or a system of objects. A new concept of linear momentum can also be used to solve physical problems, especially the problems involving collisions of objects. Linear momentum of an object whose mass is m and is moving at a velocity of v is defined as 1.Momentum is a vector quantity. 2.The heavier the object the higher the momentum 3.The higher the velocity the higher the momentum 4.Its unit is kg.m/s What can you tell from this definition about momentum? What else can use see from the definition? Do you see force? The change of momentum in a given time interval

4. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 4 Linear Momentum and Forces What can we learn from this Force-momentum relationship? Something else we can do with this relationship. What do you think it is? The relationship can be used to study the case where the mass changes as a function of time. Can you think of a few cases like this? Motion of a meteoriteTrajectory a satellite The rate of the change of particle’s momentum is the same as the net force exerted on it. When net force is 0, the particle’s linear momentum is constant. If a particle is isolated, the particle experiences no net force, therefore its momentum does not change and is conserved.

5. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 5 Conservation of Linear Momentum in a Two Particle System Consider a system with two particles that does not have any external forces exerted on it. What is the impact of Newton’s 3 rd Law? Now how would the momenta of these particles look like? If particle#1 exerts force on particle #2, there must be another force that the particle #2 exerts on #1 as the reaction force. Both the forces are internal forces and the net force in the SYSTEM is still 0. Let say that the particle #1 has momentum p 1 and #2 has p 2 at some point of time. Using momentum- force relationship And since net force of this system is 0 Therefore The total linear momentum of the system is conserved!!!

6. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 6 More on Conservation of Linear Momentum in a Two Particle System What does this mean? As in the case of energy conservation, this means that the total vector sum of all momenta in the system is the same before and after any interaction Mathematically this statement can be written as Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant. From the previous slide we’ve learned that the total momentum of the system is conserved if no external forces are exerted on the system. This can be generalized into conservation of linear momentum in many particle systems.

7. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 7 Example 9.1 Estimate an astronaut’s resulting velocity after he throws his book to a direction in the space to move to a direction. From momentum conservation, we can write vAvA vBvB Assuming the astronaut’s mass if 70kg, and the book’s mass is 1kg and using linear momentum conservation Now if the book gained a velocity of 20 m/s in +x-direction, the Astronaut’s velocity is

8. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 8 Example 9.2 A type of particle, neutral kaon (K 0 ) decays (breaks up) into a pair of particles called pions (  + and  - ) that are oppositely charged but equal mass. Assuming K 0 is initially produced at rest, prove that the two pions must have mumenta that are equal in magnitude and opposite in direction. This reaction can be written as Since K0 is produced at rest its momentum is 0. K0K0   p  p  Since this system consists of a K 0 in the initial state which results in two pions in the final state, the momentum must be conserved. So we can write Therefore, the two pions from this kaon decay have the momanta with same magnitude but in opposite direction.

9. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 9 Impulse and Linear Momentum By integrating the above equation in a time interval t i to t f, one can obtain impulse I. Impulse of the force F acting on a particle over the time interval  t=t f -t i is equal to the change of the momentum of the particle caused by that force. Impulse is the degree of which an external force changes momentum. The above statement is called the impulse-momentum theorem and is equivalent to Newton’s second law. Net force causes change of momentum  Newton’s second law So what do you think an impulse is? What are the dimension and unit of Impulse? What is the direction of an impulse vector? Defining a time-averaged forceImpulse can be rewrittenIf force is constant It is generally approximated that the impulse force exerted acts on a short time but much greater than any other forces present.

10. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 10 Example 9.3 A golf ball of mass 50g is struck by a club. The force exerted on the ball by the club varies from 0, at the instant before contact, up to some maximum value at which the ball is deformed and then back to 0 when the ball leaves the club. Assuming the ball travels 200m. Estimate the magnitude of the impulse caused by the collision. The range R of a projectile is Considering the time interval for the collision, t i and t f, initial speed and the final speed are Let’s assume that launch angle  i =45 o. Then the speed becomes: m BB vBvB Therefore the magnitude of the impulse on the ball due to the force of the club is

11. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 11 Example 9.4 In a crash test, an automobile of mass 1500kg collides with a wall. The initial and final velocities of the automobile are v i =-15.0 i m/s and v f =2.60 i m/s. if the collision lasts for 0.150 seconds, what would be the impulse caused by the collision and the average force exerted on the automobile? Let’s assume that the force involved in the collision is a lot larger than any other forces in the system during the collision. From the problem, the initial and final momentum of the automobile before and after the collision is Therefore the impulse on the automobile due to the collision is The average force exerted on the automobile during the collision is

12. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 12 Collisions Consider a case of a collision between a proton on a helium ion. The collisions of these ions never involves a physical contact because the electrostatic repulsive force between these two become great as they get closer causing a collision. Generalized collisions must cover not only the physical contact but also the collisions without physical contact such as that of electrostatic ones in a microscopic scale. t F F 12 F 21 Assuming no external forces, the force exerted on particle 1 by particle 2, F 21, changes the momentum of particle 1 is Likewise for particle 2 by particle 1 Using Newton’s 3 rd law we obtain So the momentum change of the system in the collision is 0 and the momentum is conserved

13. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 13 Example 9.5 A car of mass 1800kg stopped at a traffic light is rear-ended by a 900kg car, and the two become entangled. If the lighter car was moving at 20.0m/s before the collision what is the velocity of the entangled cars after the collision? The momenta before and after the collision are What can we learn from these equations on the direction and magnitude of the velocity before and after the collision? m1m1 20.0m/s m2m2 vfvf m1m1 m2m2 Since momentum of the system must be conserved The cars are moving in the same direction as the lighter car’s original direction to conserve momentum. The magnitude is inversely proportional to its own mass. Before collision After collision

14. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 14 Elastic and Inelastic Collisions Collisions are classified as elastic or inelastic by the conservation of kinetic energy (or momentum) before and after the collisions. A collision in which the total kinetic energy (and momentum) is the same before and after the collision. Momentum is conserved in any collisions as long as external forces negligible. Elastic Collision Two types of inelastic collisions:Perfectly inelastic and inelastic Perfectly Inelastic: Two objects stick together after the collision moving at a certain velocity together. Inelastic: Colliding objects do not stick together after the collision but some kinetic energy is lost. Inelastic Collision A collision in which the total kinetic energy is not the same before and after the collision. Note: Momentum is constant in all collisions but kinetic energy is only in elastic collisions.

15. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 15 Elastic and Perfectly Inelastic Collisions In perfectly Inelastic collisions, the objects stick together after the collision, moving together. Momentum is conserved in this collision, so the final velocity of the stuck system is How about elastic collisions? In elastic collisions, both the momentum and the kinetic energy are conserved. Therefore, the final speeds in an elastic collision can be obtained in terms of initial speeds as

16. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 16 Two dimensional Collisions In two dimension, one can use components of momentum to apply momentum conservation to solve physical problems. m2m2 m1m1 v 1i m1m1 v 1f  m2m2 v 2f  Consider a system of two particle collisions and scatters in two dimension as shown in the picture. (This is the case at fixed target accelerator experiments.) The momentum conservation tells us: And for the elastic conservation, the kinetic energy is conserved: What do you think we can learn from these relationships?

17. Feb. 25, 20021443-501 Spring 2002 Dr. J. Yu, Lecture #10 17 Example 9.9 Proton #1 with a speed 3.50x10 5 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37 o to the horizontal axis and proton #2 deflects at an angle  to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, . Since both the particles are protons m 1 =m 2 =m p. Using momentum conservation, one obtains m2m2 m1m1 v 1i m1m1 v 1f  m2m2 v 2f  Canceling m p and put in all known quantities, one obtains From kinetic energy conservation: Solving Eqs. 1-3 equations, one gets Solve this at home